# Does every vector space have a Hamel basis ? And is every linear comination representation finite?

If $$V$$ is a linear space, then a set $$B$$ of linearly independent vectors in $$V$$ that span $$V$$ is called a Hamel basis for $$V$$. Does every infinite dimensional vector space have a Hamel basis ?

My second question is : On page 55 in Erwin Kreyszig's Introductory functional analysis, it reads "Hence if B is a Hamel basis for $$V$$, then every nonzero $$v\in V$$ has a unique representation as a linear combination of (finitely many!) elements of B with nonzero scalars as coefficients."

I do not understand this because I think I have what seems to be a counterexample:

The space $$l^{\infty}$$ of all bounded sequences of real numbers has the basis $$B=\{e_i, i\geq 1\}$$, where


$$e_i$$ is the sequence with all terms zero except for the $$i$$-th term which equals 1. Now, I think of the sequence $$a=(a_i)_{i\geq 1}$$ where $$a_i=1$$. We have $$a\in l^{\infty}$$, but it does not have a linear combination representation w.r.t $$B$$ that has finitely many nonzero coefficients. What did I misss?

Thanks a lot

A Hamel basis is defined as a maximal linearly independent set. Every vector space has a Hamel basis and this well known result is proved using Zorn's Lemma. In your example $$B$$ is not a Hamel basis. (And you have proved that it is not one! You know that there is an element $$x$$ which is not a finite linear combination of members of $$B$$. This implies $$B \cup \{x\}$$ is linearly independent so $$B$$ cannot be maximal linearly independent set).

• Thank you. But, please help me understand why the set $B$ in my example is not a basis. Precisely, why are not the vectors $e_i$ linearly independent ? Is linear independence a property that only finite sets may have ?
– Medo
Nov 16, 2019 at 12:47
• @Medo I have added some details. Nov 16, 2019 at 12:50

Let us begin by defining what does it mean for an infinite set to be linearly independent:

Definition: A set $$S$$ of a vector space $$S$$ is linearly independent if for every finite collection $$(e_i)\subset S$$, if $$\sum a_ie_i =0,$$ then $$a_i = 0$$, for every $$i$$.

That is, an infinite set is linearly independent iff all of its finite subsets are linearly independently in the usual sense.

Now, we say that a set $$B$$ is a basis for $$X$$ if it is linearly independent and it generates $$X$$. This is equivalent to $$B$$ being a minimal generating set for $$X$$.

For the $$X = \ell_\infty$$ case, it is clear that the collection $$(e_i)$$ is linearly independent, but as you have showed, it does not generate $$X$$ and, therefore, it cannot be a basis. Moreover, $$X$$ does have a basis but one cannot write it explicitly for:

(1) such a basis cannot be countable (by Baire's theorem) and must have the cardinality of the continuum, making it (essentially) impossible to describe.

(2) Zorn's lemma only guarantees existence while not giving a procedure to obtain bases, as expected of something equivalent to the axiom of choice.

On certain infinite dimensional spaces, if you have a norm, you can fix this cardinality issue by allowing other notions of bases, such as Schauder bases, which allows infinite linear combinations. But not all normed spaces do have them, not even all separable ones. Schauder bases also behave very differently from Hamel bases and often one wants more properties to work with, such as unconditionality or homogeneity.

Overall, infinite dimensional spaces are a lot richer and more complicated than finite dimensional ones and Hamel bases are not a very good tool for studying them.